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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #1645 : Calculus 3

Find the gradient vector of the following function:

$f(x, y,z)=7z^2+z+x^2+2x+\ln(xy^2)$

Possible Answers:

$\left \langle 2x+2+\frac{1}{x}, \frac{1}{y}, 14z+1\right \rangle$

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 15z\right \rangle$

$\left \langle 4x+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

Correct answer:

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

Explanation:

The gradient vector of a function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=2x+2+\frac{1}{x}$ , $f_y= \frac{2}{y}$ , f_z=14z+1

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Example Question #1646 : Calculus 3

Find the equation of the plane tangent to the following function at (7, 1, 1) :

$f(x,y,z)=z^3\ln(y)+\frac{x^2}{2}$

Possible Answers:

7x+y+z=51

7x+y=51

7x+y=0

7x+y+z=0

Correct answer:

7x+y+z=51

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=x , $f_y=\frac{z^3}{y}$ , $f_z=3z^2\ln(y)$

Evaluated at the given point, the partial derivatives are

f_x=7 , f_y=1 , f_z=0

Plugging this into our equation, we get

7(x-7)+(y-1)+(z-1)=0

which simplified becomes

7x+y+z=51

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Example Question #1647 : Calculus 3

Find the gradient vector of the following function:

$f(x,y,z)=\cos(x)+\sin^2(xy)+\cos(z^2)$

Possible Answers:

$\left \langle -\sin(x)+2y\sin(xy)\cos(xy), 2x\sin(xy)\cos(xy), -2z\sin(z^2))\right \rangle$

$\left \langle \sin(x)+2y\sin(xy)\cos(xy), 2x\sin(xy)\cos(xy), 2z\sin(z^2))\right \rangle$

$\left \langle -\sin(x)+2y\sin(xy)\cos(xy), 2x\sin(xy)\cos(xy),-\sin(z^2))\right \rangle$

$\left \langle -\sin(x)+2x\sin(xy)\cos(xy), 2y\sin(xy)\cos(xy), -2z\sin(z^2))\right \rangle$

Correct answer:

$\left \langle -\sin(x)+2y\sin(xy)\cos(xy), 2x\sin(xy)\cos(xy), -2z\sin(z^2))\right \rangle$

Explanation:

The gradient vector of a function is given by

$\bigtriangledown f= \left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-\sin(x)+2y\sin(xy)\cos(xy)$ , $f_y=2x\sin(xy)\cos(xy)$ , $f_z=-2z\sin(z^2)$

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Example Question #111 : Applications Of Partial Derivatives

Find the equation of the tangent plane at the point (1,1,-2) to the surface

x^3+y^2+3z^2=5

Possible Answers:

3x+2y-12z-29=0

2x+3y+12z+19=0

3x-2y+12z+29=0

2x+3y+12z-18=0

Correct answer:

3x+2y-12z-29=0

Explanation:

The equation of the surface given in the problem is the level surface (with k=5 ) of the function F(x,y,z)=x^3+y^2+3z^2 . Hence, we can apply the definition of the tangent plane to a level surface to calculate the plane passing through the point (1,1,-2) at an orthogonal direction (that is, perpendicularly) to the point's normal vector.

Recall this definition of the tangent plane to a level surface F(x,y,z) :

F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0

We have F(x,y,z)=x^3+y^2+3z^2 and the point (1,1,-2) . Calculating the partial derivatives of as well as the values of each of these derivatives at yields the following information:

F_x(x,y,z)=3x^2 ,

F_y(x,y,z)=2y ,

F_z(x,y,z)=6z ,

F_x(1,1,-2)=3(1)^2=3 ,

F_y(1,1,-2)=2(1)=2 ,

F_z(1,1,-2)=6(-2)=-12 .

Therefore, the aforementioned definition of the tangent plane gives the equation for the tangent plane of F(x,y,z) at (1,1,2) as

3(x-1)+2(y-1)-12(z+2)=0

$\Rightarrow 3x-3+2y-2-12z-24=0$

$\Rightarrow 3x+2y-12z-29=0$

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Example Question #1648 : Calculus 3

Find the equation of the tangent plane to the following function at (0, 3, 2) :

$f(x,y,z)=xy\sin(x)+z$

Possible Answers:

$z=\frac{1}{2}$

z=2

z=0

x+y+z=2

Correct answer:

z=2

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=y(\sin(x)+x\cos(x))$

$f_y=x\sin(x)$

f_z=1

The partial derivatives evaluated at the given point are

f_x=0 , f_y=0 , f_z=1

Plugging this into the equation above, we get

1(z-2)=0

which simplifies to

z=2

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Example Question #91 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector of the function:

$f(x,y,z)=xe^{z^2}+y\ln(z^2)$

Possible Answers:

$\left \langle 0, 0, 2xze^{z^2}+\frac{2y}{z}\right \rangle$

$\left \langle e^{z^2},\ln(z^2), 2xze^{z^2}+\frac{2}{z}\right \rangle$

$\left \langle e^{z^2},\ln(z^2), 2xze^{z^2}+\frac{2y}{z}\right \rangle$

$\left \langle e^{z^2},\ln(z^2), 2xze^{z^2}+\frac{2y}{z^2}\right \rangle$

Correct answer:

$\left \langle e^{z^2},\ln(z^2), 2xze^{z^2}+\frac{2y}{z}\right \rangle$

Explanation:

The gradient vector is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=e^{z^2}$ , $f_y=\ln(z^2)$ , $f_z= 2xze^{z^2}+\frac{2y}{z}$

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Example Question #91 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector for the following function:

f(x,y,z)=x^3y^3+x^2z^4e^z

Possible Answers:

$\left \langle 3x^2y^3+2xz^4e^z, 3x^2y^2, x^2e^z(4z^3+z^4)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, x^2z^2e^z(6z^2)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, e^z(4x^2z^3+z^4)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, x^2e^z(4z^3+z^4)\right \rangle$

Correct answer:

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, x^2e^z(4z^3+z^4)\right \rangle$

Explanation:

The gradient vector of the function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=3x^2y^3+2xz^4e^z , f_y=3x^3y^2 , f_z=x^2e^z(4z^3+z^4)

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Example Question #92 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector for the following function:

$f(x,y,z)=z^3+\cos(z)+y\ln(y)$

Possible Answers:

$\left \langle 0, \ln(y)+1, 3z^2+\sin(z)\right \rangle$

$\left \langle 0, \frac{1}{y}, 3z^2-\sin(z)\right \rangle$

$\left \langle 0, \ln(y)+\frac{1}{y}, 3z^2-\sin(z)\right \rangle$

$\left \langle 0, y+1, 3z^2-\sin(z)\right \rangle$

$\left \langle 0, \ln(y)+1, 3z^2-\sin(z)\right \rangle$

Correct answer:

$\left \langle 0, \ln(y)+1, 3z^2-\sin(z)\right \rangle$

Explanation:

The gradient vector of the function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=0 , $f_y=\ln(y)+1$ , $f_z=3z^2-\sin(z)$

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Example Question #1652 : Calculus 3

Write the equation of the tangent plane to the following function at (4, 2, 1) :

f(x,y,z)=x^2+y^2+8z^2

Possible Answers:

2x+y+4z=14

4x+2y+z=14

x+y+z=56

2x+y+4z=0

Correct answer:

2x+y+4z=14

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=2x , f_y=2y , f_z=16z

The partial derivatives evaluated at the given point

f_x=8 , f_y=4 , f_z=16

Plugging our known information into the equation above, we get

8(x-4)+4(y-2)+16(z-1)=0

which simplifies to

2x+y+4z=14

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Example Question #94 : Gradient Vector, Tangent Planes, And Normal Lines

Find the function with the gradient:

$\vec{\triangledown}f=2xz\hat{i}+z\hat{j}+\left(x^2+yz+\frac{1}{z}\right)\hat{k}$

that satisfies the condition f(1,0,1)=9

Possible Answers:

$f(x,y,z)=zx^2+\ln(z)+xy^2+2$

$f(x,y,z)=z^2x+\ln(z)+\frac{1}{2}zx^2$

f(x,y,z)=zx^2+zy+ln(z)+8

$f(x,y,z)=zx^2+zy+\ln(z)+4$

$f(x,y,z)=zx^2+zy^2+\ln(z)+4$

Correct answer:

f(x,y,z)=zx^2+zy+ln(z)+8

Explanation:

$\vec{\triangledown}f=2xz\hat{i}+z\hat{j}+\left(x^2+y+\frac{1}{z}\right)\hat{k}$

One easy solution is to just compute the gradients of each of the multiple choices for this problem and then apply the given condition f(1,0,1)=9 . If your exam is not multiple choice, you will have to use a more insightful approach. Compare the given gradient to the general definition of a gradient:

$\vec{\triangledown}f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$ (1)

Since we obtain the gradient by computing the partial derivatives of each independent variable, we can obtain the function by integrating each component and then combining the results to write a single expression for

$f=\int \frac{\partial f}{\partial x}dx=\int 2xz dx = x^2z+C_1$

$f=\int \frac{\partial f}{\partial y}dy=\int z dy= zy+C_2$

$f=\int \frac{\partial f}{\partial z}dz=\int \left(x^2+y+\frac{1}{z} \right )dz = x^2z+yz+\ln(z)+C_3$

Looking at the results, note that each we obtained may include either some of the terms or all of the terms of the functions actual function. In this case, integrating the $\hat{k}$ component gave back all the terms of .

So our function will have the form:

f(x,y,z)=zx^2+zy+ln(z)+C

Apply the condition f(1,0,1)=9 :

$f(1,0,1)=1+\ln(1)+C = 9$

C=8

Therefore the function we were looking for has the form:

f(x,y,z)=zx^2+zy+ln(z)+8

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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1645 : Calculus 3

Example Question #1646 : Calculus 3

Example Question #1647 : Calculus 3

Example Question #111 : Applications Of Partial Derivatives

Example Question #1648 : Calculus 3

Example Question #91 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #91 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #92 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1652 : Calculus 3

Example Question #94 : Gradient Vector, Tangent Planes, And Normal Lines

All Calculus 3 Resources

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